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Unformatted text preview: ~ E2) where ~ is the mean logarithmic energy lost by a neutron in an elastic scatter with a nucleus of atomic mass number A, namely a f. = 1 + :;:lna with a == ~=!2: (A + 1)2 u For sodium (A = 23), we find a = 0.8403 and f. = 0.08448. Then the number of elastic scatters with carbon required to reduce a neutron's energy from 2 MeV to 0.025 eV is 1 ( El ) 1 ( 2 X 106 ) n = ~ In "E; = 0.08448 In 0.025 = 215. July 24, 2002...
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This note was uploaded on 02/01/2010 for the course ECE 4130 taught by Professor Cady during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 CADY

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